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प्रश्न
How does Avogadro's law explain Gay - lussac's law of combining volumes?
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उत्तर
Gay-Lussac experimentally determined that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas.
According to Avogadro's law, if :
1 volume of hydrogen contains n molecules of the gas then 1 volume of chlorine also contains n molecules of the gas. Therefore 2 volume of hydrogen chloride contain 2n molecules of the gas.
\[\ce{H2 + Cl2 → 2HCI}\]
1 vol 1 vol 2 vol (by Gay-Lussac)
n n 2n (By Avogadro)
but hydrogen and chlorine are diatomic.
So, 2 atoms + 2 atoms → 2 molecules
1 atom + 1 atom → 1 molecule
i.e. 1 molecule of hydrogen chloride is formed when 1 atom of hydrogen combines with 1 atom of chlorine. Thus Avogadro's law explains Gay-Lussac's law of combining volumes.
संबंधित प्रश्न
State Gay-Lussac’s law of combining volumes.
Water can split into hydrogen and oxygen under suitable conditions. The equations representing the change is: 2H2O(I) → 2H2 (g) + O2(g)
If a given experiment results in 2500cm3 of hydrogen being produced, what volume of oxygen is liberated at the same time under the same conditions of temperature and pressure?
When gases react their volumes bear a simple ratio to each other, under the same conditions of temperature and pressure. Who proposed this gas law?
What volume of oxygen would be required for complete combustion of 100L of ethane according to the following equation?
2C2H6 + 7O2 → 4CO2 + 6H2O
The reaction 4N2O + CH4 → CO2 + 2H2O + 4N2 takes place in the gaseous state. If the volumes of all the gases are measured at the same temperature, and pressure, calculate the volume of dinitrogen oxide (N2O), required to give 150cm3 of steam.
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is `1/5`th of air)
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.
- Which sample of gas contains the maximum number of molecules?
- If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
- If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
- Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate composition of the resulting mixture.
1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen formed:
\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
